我有以下两个文件。
如何使用 AJAX 填充标签和值
例如,如果值为伊利诺伊州芝加哥。提交表格时如何获得仅芝加哥的值?
下面的字段填充为,例如,伊利诺伊州芝加哥
<!DOCTYPE html>
<html>
<head>
<title></title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/bootstrap-3-typeahead/4.0.2/bootstrap3-typeahead.min.js"></script>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" />
</head>
<body>
<br /><br />
<div class="container" style="width:600px;">
<h2 align="center"></h2>
<br /><br />
<label>Search Country</label>
<input type="text" name="city" id="city" class="form-control input-lg" autocomplete="off" placeholder="Type City Name" />
</div>
</body>
</html>
<script>
$(document).ready(function(){
$('#city').typeahead({
source: function(query, result)
{
$.ajax({
url:"TypeaheadTest2.php",
method:"POST",
data:{query:query},
dataType:"json",
success:function(data)
{
result($.map(data, function(item){
return item;
}));
}
})
}
});
});
</script>
下面是文件 TypeaheadTest2.php
<?php
$connect = mysqli_connect("localhost", "", "", "");
$request = mysqli_real_escape_string($connect, $_POST["query"]);
$query = "
SELECT * FROM state WHERE state LIKE '".$request."%' OR city LIKE '%".$request."%'
";
$result = mysqli_query($connect, $query);
$data = array();
if(mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_assoc($result))
{
$data[] = $row['city'] . ' ' . $row['state'];
}
echo json_encode($data);
}
?>
最佳答案
如果我的理解是正确的,那么您已经非常接近您的期望了。
我发现您正在服务器端连接城市和州名称。但您只想在预输入选项上显示城市名称。
我建议您将城市名称和州名称作为 PHP 对象的单独属性传递,然后在 Javascript 端根据需要使用它们。
这是示例代码, PHP:
<?php
$connect = mysqli_connect("localhost", "", "", "");
$request = mysqli_real_escape_string($connect, $_POST["query"]);
$query = "
SELECT * FROM state WHERE state LIKE '".$request."%' OR city LIKE '%".$request."%'
";
$result = mysqli_query($connect, $query);
$data = array();
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_assoc($result)){
$data[] = array('city'=>$row['city'], 'state'=>$row['state']);
}
echo json_encode($data);
}
?>
Javascript:
$(document).ready(function(){
$('#city').typeahead({
source: function(query, result) {
$.ajax({
url:"TypeaheadTest2.php",
method:"POST",
data:{query:query},
dataType:"json",
success:function(data) {
result($.map(data, function(item){
return item.city;
}));
}
});
}
});
});
关于php - AJAX:如何使字段的标签和值不同?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45871364/